Question

The area between the curves $y = \tan x,y = \cot x$ and x-axis in the interval $\left[ {0,\dfrac{\pi }{2}} \right]$ is
A.log2
B.log3
C.log$\sqrt 2$
D. None of these

Answer 1

The questions which are related to intervals first draw a diagram for solving and understanding the meaning of the question. Using the values of trigonometry function we can see that $\tan \theta$ and $\cot \theta$ are used in the question so values of the tan and cot will help us to solve the question correctly. In this question also use the property of log which is $\log |_n^m = \log m - \log n$.

Complete step-by-step answer:

In the diagram point A = $\dfrac{\pi }{4}$ and point B =$\dfrac{\pi }{2}$
As the question mention we have to find the area between the intervals $(0,\dfrac{\pi }{2})$
In the diagram the bisecting point is $\dfrac{\pi }{4}$ because the value of $\tan \dfrac{\pi }{4}$ and $\cot \dfrac{\pi }{4}$ are equal to 1.
In the diagram first the value of tan is between $\dfrac{\pi }{4}$ and 1 after that its infinite
The value of cot in first is $\dfrac{\pi }{2}$ and after that its infinite
So, the area between the intervals$(0,\dfrac{\pi }{2})$is
$= \int\limits_0^{\dfrac{\pi }{4}} {\tan xdx + \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}} {\cot xdx} }$
Now derivative of $\tan x = \sec x$and$\cot x = \sin x$
$= \log \sec x|_0^{\dfrac{\pi }{4}} + \log \sin x|_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}$
Now put the values of $\sec \dfrac{\pi }{4} = \sqrt 2 ,\sec 0 = 1,\sin \dfrac{\pi }{2} = 1,\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$
$= \log \sqrt 2 - \log 1 + \log 1 - \log \dfrac{1}{{\sqrt 2 }}$
Now $\log 1$ and $- \log 1$ got cancel by each other
We know that we can write $\log 2 = \dfrac{1}{2}\log 2$ and $\log \dfrac{1}{{\sqrt 2 }} = \dfrac{1}{2}\log 2$
= $\dfrac{1}{2}\log 2$ $+$ $\dfrac{1}{2}\log 2$ $=$\log 2

So, the correct option is A.

Note: For solving questions related to log always remember the properties of log and values of trigonometry. Students make mistakes while taking the values and finding the derivatives. You don’t need the derivatives in detail in this question, just simply learn the basic derivatives of the trigonometric function and directly put it in the equation. Always cancel the same values of opposite signs like plus and minus.